Methanol (CH3OH) is used as fuel in race cars and is a potential replacement for gasoline.  It is synthesized by a combination of hydrogen and carbon monoxide.  Suppose 65.8 g of carbon monoxide...

Methanol (CH3OH) is used as fuel in race cars and is a potential replacement for gasoline.  It is synthesized by a combination of hydrogen and carbon monoxide.  Suppose 65.8 g of carbon monoxide reacts with 8.6 g of hydrogen.

What is the limiting reagent? 

Expert Answers
ncchemist eNotes educator| Certified Educator

First, let's look at the balanced chemical equation for the synthesis of methanol from carbon monoxide (CO) and hydrogen gas (H2).

CO + 2H2 --> CH3OH

Notice that we need two molecules of hydrogen gas for every molecule of CO to make a molecule of methanol.  Now let's take the mass of each reagent and convert them to moles.  We do this by dividing by the molecular weight for each compound.

65.8 g CO * (mole/44.01 g) = 1.50 moles CO

8.6 g H2 * (mole/2.02 g) = 4.26 moles H2

From the above equation we know that 2 moles of H2 are required for every mole of CO.  Since we have 1.5 moles of CO, that would require 1.50*2=3 moles of H2 to completely consume it.  We have 4.26 moles of H2 in actuality, so we will consume all of the CO present and still have 4.26-3=1.26 moles of H2 remaining.  Since all of the CO will be consumed and H2 will remain, that makes CO the limiting reagent.

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